NCERT Exemplar Solutions Class 10 Science Metals and Non-metals — Short and Long Answer Questions Q37–65

NCERT Exemplar Problems, Class 10 Science — Metals and Non-metals, Short Answer Questions 37 to 58 and Long Answer Questions 59 to 65. All 29 questions solved.

This is the longest band in the Class 10 Science exemplar, and it is dominated by one kind of question: an element is described by what it does, and you have to name it. Thirteen of these twenty-nine are of that form. They are quick once you know where to look, and the useful habit is to find the clue that fixes the element before worrying about the compound — “Group 2”, “molecular mass 40”, “largest constituent of air”, “used in the thermite process” each pin down a single element in one step.

The rest of the band is metallurgy, and the whole of it fits in one framework:

Reactivity	Extraction route
High (K, Na, Ca, Mg, Al)	Electrolysis of the molten salt
Middle (Zn, Fe, Pb, Cu)	Ore → oxide by roasting (sulphide) or calcination (carbonate) → reduce with carbon
Low (Hg, Ag, Au)	Often found native; sulphide ore gives the metal on heating alone

Roasting is heating a sulphide ore strongly in air; calcination is heating a carbonate ore in the absence of air. Both give the oxide, and the reason both are done is Question 39: an oxide is much easier to reduce than a sulphide or a carbonate.

Key insight. Every extraction question in this band is answered by asking two things in order — where in the reactivity series does this metal sit? and is the ore a sulphide or a carbonate? The first chooses the route, the second chooses whether the first step is roasting or calcination. Nothing else is needed.

Short Answer Questions

Question 37

Iqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in reaction mixture. He made the same observations when this element was treated with hydrochloric acid. Suggest how can he identify the produced gas. Write chemical equations for both the reactions.

Solution. The description contains one very unusual property: the element gives a gas with both an alkali and an acid. Only an amphoteric metal does that, and at this level that means aluminium or zinc. The word divalent decides between them — zinc forms $\mathrm{Zn^{2+}}$, aluminium $\mathrm{Al^{3+}}$ — so M is zinc.

The gas in both cases is hydrogen, and it is identified by bringing a burning splint to the mouth of the tube: the gas burns with a squeaky pop sound.

$$\mathrm{Zn} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2ZnO_2} + \mathrm{H_2}$$

$$\mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl_2} + \mathrm{H_2}$$

Written in the general form the key uses, with M for the metal, these are $\mathrm{M} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2MO_2} + \mathrm{H_2}$ and $\mathrm{M} + 2\mathrm{HCl} \rightarrow \mathrm{MCl_2} + \mathrm{H_2}$ — and writing them that way makes the point that the answer follows from divalent and amphoteric without needing the element’s name at all.

Question 38

During extraction of metals, electolytic refining is used to obtain pure metals. (a) Which material will be used as anode and cathode for refining of silver metal by this process? (b) Suggest a suitable electrolyte also. (c) In this electrolytic cell, where do we get pure silver after passing electric current?

Solution. Electrolytic refining always has the same three parts, whatever the metal, and silver is no exception.

(a) The anode is the impure metal — impure silver — and the cathode is a thin strip of the pure metal, pure silver. The anode is the one that dissolves, so it has to be the impure block; the cathode is the one that grows, so it has to be clean to start with.

(b) The electrolyte must be a soluble salt of the same metal, so that the ions arriving at the cathode are silver ions. Silver nitrate solution, $\mathrm{AgNO_3}$, acidified, is the usual choice.

(c) Pure silver is deposited on the cathode. The anode dissolves as $\mathrm{Ag^+}$ ions, which travel to the negative cathode and are discharged there as the metal:

$$\text{At the anode: } \mathrm{Ag}(s) \rightarrow \mathrm{Ag^+}(aq) + e^-$$

$$\text{At the cathode: } \mathrm{Ag^+}(aq) + e^- \rightarrow \mathrm{Ag}(s)$$

The impurities, being either insoluble or less reactive, do not dissolve and collect under the anode as anode mud.

Question 39

Why should the metal sulphides and carbonates be converted to metal oxides in the process of extraction of metal from them?

Solution. Because a metal oxide is far easier to reduce than a sulphide or a carbonate, and reduction is the step that actually produces the metal.

Once the ore has been converted to the oxide, a single cheap reducing agent — carbon, as coke — will pull the oxygen off:

$$\mathrm{ZnO} + \mathrm{C} \xrightarrow{\text{Heat}} \mathrm{Zn} + \mathrm{CO}$$

There is no correspondingly simple, cheap way to strip sulphur from a sulphide or a carbonate group from a carbonate directly. Carbon has a strong affinity for oxygen and readily forms carbon monoxide or dioxide, which escape as gases and drive the reaction forward; it has no such affinity for sulphur.

So the conversion is a preparation step, and it is done in two ways depending on the ore:

• Roasting — a sulphide ore heated strongly in the presence of air: $2\mathrm{ZnS} + 3\mathrm{O_2} \rightarrow 2\mathrm{ZnO} + 2\mathrm{SO_2}$
• Calcination — a carbonate ore heated in the absence of air: $\mathrm{ZnCO_3} \rightarrow \mathrm{ZnO} + \mathrm{CO_2}$

Both end at the same place, the oxide, from which carbon reduction gives the metal.

Question 40

Generally, when metals are treated with mineral acids, hydrogen gas is liberated but when metals (except Mn and Mg), treated with $\mathrm{HNO_3}$, hydrogen is not liberated, why?

Solution. Because nitric acid is not only an acid but a strong oxidising agent.

The displacement itself happens as usual — a metal above hydrogen in the reactivity series does push hydrogen out of the acid. But the hydrogen produced does not survive to escape. The nitric acid oxidises it at once to water:

$$\mathrm{H_2} + [\mathrm{O}] \rightarrow \mathrm{H_2O}$$

and the nitric acid is itself reduced, coming off as an oxide of nitrogen instead. Which oxide depends on the concentration and the metal: dinitrogen oxide $\mathrm{N_2O}$ with very dilute acid, nitrogen monoxide $\mathrm{NO}$ with dilute acid, nitrogen dioxide $\mathrm{NO_2}$ with concentrated acid. A worked example is $4\mathrm{Zn} + 10\mathrm{HNO_3} \rightarrow 4\mathrm{Zn(NO_3)_2} + 5\mathrm{H_2O} + \mathrm{N_2O}$.

The exception. Manganese and magnesium do give hydrogen, but only with very dilute nitric acid. There the acid’s oxidising power is weakest and these two metals react fast enough that the hydrogen escapes before it can be oxidised.

Question 41

Compound X and aluminium are used to join railway tracks. (a) Identify the compound X (b) Name the reaction (c) Write down its reaction.

Solution.

(a) X is ferric oxide, $\mathrm{Fe_2O_3}$ — iron(III) oxide.

(b) The reaction is the thermite reaction.

(c) Aluminium is more reactive than iron, so it displaces the iron from its oxide:

$$\mathrm{Fe_2O_3}(s) + 2\mathrm{Al}(s) \rightarrow 2\mathrm{Fe}(l) + \mathrm{Al_2O_3}(s) + \text{Heat}$$

What makes the reaction useful is that it is enormously exothermic — so much heat is released that the iron is produced as a molten liquid, at well over $2000^\circ\mathrm{C}$. That molten iron is poured straight into the gap between two rail ends, where it solidifies and welds them together. The state symbol $(l)$ on the iron is therefore not a detail; it is the entire point of the process.

The same reaction is used to weld cracked machine parts in situ, where taking the part to a furnace is impossible.

Question 42

When a metal X is treated with cold water, it gives a basic salt Y with molecular formula XOH (Molecular mass = 40) and liberates a gas Z which easily catches fire. Identify X, Y and Z and also write the reaction involved.

Solution. The molecular mass is the clue that does all the work, so use it first.

Y has the formula $\mathrm{XOH}$ with molecular mass 40. The $\mathrm{OH}$ group contributes $16 + 1 = 17$, so

$$\text{atomic mass of X} = 40 – 17 = 23$$

and the element of atomic mass 23 is sodium. So $\mathrm{X} = \mathrm{Na}$ and $\mathrm{Y} = \mathrm{NaOH}$.

Every clue then checks out. Sodium reacts with cold water, as the top band of the reactivity series does. Sodium hydroxide is a strong base, which matches “basic salt”. And the gas is hydrogen, which “easily catches fire” — in fact the heat of this reaction is enough to ignite it without any flame being applied, which is why sodium in water is dangerous:

$$2\mathrm{Na}(s) + 2\mathrm{H_2O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{H_2}(g) + \text{Heat}$$

Note that the formula $\mathrm{XOH}$ with a single hydroxide group already rules out calcium and magnesium, which are divalent and would give $\mathrm{X(OH)_2}$. The molecular mass then chooses between sodium and potassium.

Question 43

A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.

Solution. Two forms of the same element with wildly different properties are allotropes, and the pair described here is the most famous in chemistry.

Y is the hardest natural substance — that is diamond, and nothing else competes for the description.

Z is a good conductor of electricity, which is remarkable for a non-metal — that is graphite, from Question 32.

Both are forms of carbon, so $\mathrm{X} = \mathrm{C}$.

The contrast is entirely structural, which is worth saying because it is the real content of the question. In diamond each carbon atom is bonded to four others in a rigid three-dimensional network; every electron is committed to a bond, so it is extremely hard and does not conduct. In graphite each carbon is bonded to only three others in flat sheets; the sheets slide over one another, making graphite soft and slippery, and the fourth electron per atom is free to move, making it conduct.

Same atoms, different arrangement, opposite properties.

Question 44

The following reaction takes place when aluminium powder is heated with $\mathrm{MnO_2}$

$$3\mathrm{MnO_2}(s) + 4\mathrm{Al}(s) \rightarrow 3\mathrm{Mn}(l) + 2\mathrm{Al_2O_3}(l) + \text{Heat}$$

(a) Is aluminium getting reduced? (b) Is $\mathrm{MnO_2}$ getting oxidised?

Solution. Both answers are no, and the question is testing whether the definitions are held the right way round rather than anything about this particular reaction.

(a) Aluminium starts as the free metal and ends up in $\mathrm{Al_2O_3}$ — it has gained oxygen. Gaining oxygen is oxidation, so aluminium is being oxidised, not reduced. Aluminium is the reducing agent here.

(b) Manganese dioxide starts with two oxygen atoms per manganese and ends as free manganese — it has lost oxygen. Losing oxygen is reduction, so $\mathrm{MnO_2}$ is being reduced, not oxidised. $\mathrm{MnO_2}$ is the oxidising agent.

The pattern is the same one as the thermite reaction of Question 41 — aluminium displacing a less reactive metal from its oxide, releasing enough heat to leave both products molten, as the $(l)$ symbols show.

A useful check on any answer of this kind: the two reactants must go in opposite directions. If you find yourself saying both are oxidised or both are reduced, something has gone wrong.

Question 45

What are the constituents of solder alloy? Which property of solder makes it suitable for welding electrical wires?

Solution. Solder is an alloy of lead and tin, in roughly equal parts.

The property that matters is its low melting point — around $200^\circ\mathrm{C}$, far below that of either the copper wire being joined or of lead and tin taken separately. That means it can be melted with a small hand-held iron and flowed into a joint without heating the wires enough to damage them or their insulation. It solidifies quickly on cooling, and being a metal it conducts, so the joint carries current as well as holding the wires together.

This is a good illustration of why alloys are made at all: an alloy can have a melting point lower than either of its constituents, which is not something you could guess from the components. Lead melts at $327^\circ\mathrm{C}$ and tin at $232^\circ\mathrm{C}$, yet the mixture melts below both.

Question 46

A metal A, which is used in thermite process, when heated with oxygen gives an oxide B, which is amphoteric in nature. Identify A and B. Write down the reactions of oxide B with HCl and NaOH.

Solution. The metal used in the thermite process is aluminium — Question 41 — so $\mathrm{A} = \mathrm{Al}$, and the second clue confirms it, because aluminium oxide is the standard amphoteric oxide of Question 31. So $\mathrm{B} = \mathrm{Al_2O_3}$.

$$4\mathrm{Al} + 3\mathrm{O_2} \rightarrow 2\mathrm{Al_2O_3}$$

Being amphoteric means B reacts with acids and with bases, giving a salt and water in both cases.

With an acid, it behaves as a basic oxide:

$$\mathrm{Al_2O_3} + 6\mathrm{HCl} \rightarrow 2\mathrm{AlCl_3} + 3\mathrm{H_2O}$$

With a base, it behaves as an acidic oxide:

$$\mathrm{Al_2O_3} + 2\mathrm{NaOH} \rightarrow 2\mathrm{NaAlO_2} + \mathrm{H_2O}$$

The product of the second is sodium aluminate. Check that balance carefully — it is the one students most often get wrong: two aluminiums need two $\mathrm{NaAlO_2}$, which need two sodiums, hence $2\mathrm{NaOH}$, and the two hydrogens then make exactly one water.

Question 47

A metal that exists as a liquid at room temperature is obtained by heating its sulphide in the presence of air. Identify the metal and its ore and give the reaction involved.

Solution. The metal that is liquid at room temperature is mercury — Question 17 — and its sulphide ore is cinnabar, $\mathrm{HgS}$.

Mercury sits at the very bottom of the reactivity series, and that is what makes its extraction so simple. Heating the sulphide in air first roasts it to the oxide:

$$2\mathrm{HgS} + 3\mathrm{O_2} \xrightarrow{\text{Heat}} 2\mathrm{HgO} + 2\mathrm{SO_2}$$

and then continued heating decomposes that oxide straight to the metal, with no reducing agent needed at all:

$$2\mathrm{HgO} \xrightarrow{\text{Heat}} 2\mathrm{Hg} + \mathrm{O_2}$$

The second step is the striking one. For zinc or iron you would have to add carbon at this point; mercury oxide is so unstable that heat alone is enough. That is the general rule for the low-reactivity band — their oxides decompose on heating — and it is the left-hand branch of the flowchart in Question 60.

Question 48

Give the formulae of the stable binary compounds that would be formed by the combination of following pairs of elements.

(a) Mg and $\mathrm{N_2}$    (b) Li and $\mathrm{O_2}$    (c) Al and $\mathrm{Cl_2}$    (d) K and $\mathrm{O_2}$

Solution. Every one of these is a metal with a non-metal, so every compound is ionic, and the formula follows from the charges: work out the ion each element forms, then combine them in whatever ratio makes the total charge zero.

Element	Outer electrons	Ion formed
Mg	2	$\mathrm{Mg^{2+}}$
Li, K	1	$\mathrm{Li^+}$, $\mathrm{K^+}$
Al	3	$\mathrm{Al^{3+}}$
N	5, needs 3	$\mathrm{N^{3-}}$
O	6, needs 2	$\mathrm{O^{2-}}$
Cl	7, needs 1	$\mathrm{Cl^-}$

(a) $\mathrm{Mg^{2+}}$ with $\mathrm{N^{3-}}$: three magnesiums give $+6$, two nitrides give $-6$ — $\mathbf{Mg_3N_2}$.

(b) $\mathrm{Li^+}$ with $\mathrm{O^{2-}}$: two lithiums to one oxide — $\mathbf{Li_2O}$.

(c) $\mathrm{Al^{3+}}$ with $\mathrm{Cl^-}$: one aluminium to three chlorides — $\mathbf{AlCl_3}$.

(d) $\mathrm{K^+}$ with $\mathrm{O^{2-}}$: two potassiums to one oxide — $\mathbf{K_2O}$.

Note that the subscripts in $\mathrm{N_2}$, $\mathrm{O_2}$ and $\mathrm{Cl_2}$ in the question describe the element as it exists, as a diatomic molecule, and have nothing to do with the formula of the compound. Carrying them across is the standard error here.

Question 49

What happens when

(a) $\mathrm{ZnCO_3}$ is heated in the absence of oxygen? (b) a mixture of $\mathrm{Cu_2O}$ and $\mathrm{Cu_2S}$ is heated?

Solution.

(a) Heating a carbonate ore in the absence of air is calcination, and it decomposes the carbonate to the oxide:

$$\mathrm{ZnCO_3}(s) \xrightarrow{\text{Heat}} \mathrm{ZnO}(s) + \mathrm{CO_2}(g)$$

The words “in the absence of oxygen” in the question are what make it calcination rather than roasting. Zinc carbonate is the ore calamine, and this is the first step of extracting zinc from it.

(b) This is auto-reduction — also called self-reduction — and it is one of the neatest reactions in the chapter. The oxide and the sulphide reduce each other, with no external reducing agent at all:

$$2\mathrm{Cu_2O} + \mathrm{Cu_2S} \xrightarrow{\text{Heat}} 6\mathrm{Cu} + \mathrm{SO_2}$$

The oxygen of the $\mathrm{Cu_2O}$ combines with the sulphur of the $\mathrm{Cu_2S}$ and leaves as sulphur dioxide, and all six coppers are set free as the metal.

Check the balance: six coppers on each side, two oxygens on the left and two in $\mathrm{SO_2}$, one sulphur each side. This is the final step of copper extraction, and Question 62 asks you to place it in the full sequence.

Question 50

A non-metal A is an important constituent of our food and forms two oxides B and C. Oxide B is toxic whereas C causes global warming

(a) Identify A, B and C (b) To which Group of Periodic Table does A belong?

Solution.

(a) Two oxides, one toxic and one a greenhouse gas, is a description that fits carbon and nothing else. So $\mathrm{A} = \mathrm{C}$, carbon — and it is indeed an important constituent of food, since carbohydrates, proteins and fats are all carbon compounds.

$\mathrm{B}$ is carbon monoxide, CO, the toxic one. It is toxic because it binds to haemoglobin far more strongly than oxygen does, so the blood can no longer carry oxygen. It is produced by incomplete combustion, which is why a stove burning in a poorly ventilated room is dangerous.

$\mathrm{C}$ is carbon dioxide, $\mathrm{CO_2}$, the greenhouse gas. It is not toxic at ordinary concentrations — we exhale it — but it traps infrared radiation in the atmosphere and is the principal driver of global warming.

(b) Carbon has four electrons in its outer shell and belongs to Group 14 of the periodic table.

Question 51

Give two examples each of the metals that are good conductors and poor conductors of heat respectively.

Solution. All metals conduct heat better than non-metals do, so “poor conductor” here means poor relative to other metals.

Good conductors of heat: silver and copper. Silver is the best thermal conductor of all the elements and copper is close behind, which is why cooking vessels are made of copper or aluminium and why copper is used for heat sinks.

Poor conductors of heat: lead and mercury. Both conduct heat far worse than silver — lead at roughly one-twelfth of silver’s conductivity and mercury worse still.

The reason is the same mechanism that carries electric current. In a metal, heat is transported mainly by the free electrons, so a metal that conducts electricity well generally conducts heat well too. Silver and copper are the best electrical conductors and the best thermal ones; lead and mercury are poor at both.

Question 52

Name one metal and one non-metal that exist in liquid state at room temperature. Also name two metals having melting point less than 310 K ($37^\circ\mathrm{C}$)

Solution. Only two elements are liquid at room temperature, and one is a metal and one a non-metal, which is what makes them worth learning as a pair.

The metal is mercury (Hg), freezing at about $-39^\circ\mathrm{C}$. The non-metal is bromine (Br), a dark red-brown liquid freezing at about $-7^\circ\mathrm{C}$.

For the second part, note that $310\ \mathrm{K}$ is $37^\circ\mathrm{C}$ — body temperature — so the question is asking for metals that would melt in your hand. Two do: caesium (Cs), which melts at $302\ \mathrm{K}$, and gallium (Ga), which melts at $303\ \mathrm{K}$. A lump of gallium held in the palm turns to liquid, which is a well-known demonstration.

Neither counts as “liquid at room temperature”, because room temperature is taken as about $25^\circ\mathrm{C}$ or $298\ \mathrm{K}$ and both are solid there. The question separates the two categories deliberately.

Question 53

An element A reacts with water to form a compound B which is used in white washing. The compound B on heating forms an oxide C which on treatment with water gives back B. Identify A, B and C and give the reactions involved.

Solution. The clue that fixes the element is used in white washing: that is slaked lime, calcium hydroxide. So $\mathrm{B} = \mathrm{Ca(OH)_2}$, and therefore $\mathrm{A} = \mathrm{Ca}$, calcium.

Calcium is in the cold-water band of the reactivity series, so the first reaction works:

$$\mathrm{Ca}(s) + 2\mathrm{H_2O}(l) \rightarrow \mathrm{Ca(OH)_2}(aq) + \mathrm{H_2}(g)$$

Heating calcium hydroxide drives off water and leaves the oxide, so $\mathrm{C} = \mathrm{CaO}$, quicklime:

$$\mathrm{Ca(OH)_2} \xrightarrow{\text{Heat}} \mathrm{CaO} + \mathrm{H_2O}$$

and adding water to quicklime gives the hydroxide back, which is slaking of lime:

$$\mathrm{CaO} + \mathrm{H_2O} \rightarrow \mathrm{Ca(OH)_2}$$

The last two reactions are each other’s reverse, which is the “gives back B” of the question and is a useful structural clue in its own right: an oxide and a hydroxide of the same metal are always related this way.

The whitewashing itself involves a third reaction worth knowing — the slaked lime on the wall slowly absorbs carbon dioxide from the air and sets to hard calcium carbonate, $\mathrm{Ca(OH)_2} + \mathrm{CO_2} \rightarrow \mathrm{CaCO_3} + \mathrm{H_2O}$, which is what gives the shiny white finish.

Question 54

An alkali metal A gives a compound B (molecular mass = 40) on reacting with water. The compound B gives a soluble compound C on treatment with aluminium oxide. Identify A, B and C and give the reaction involved.

Solution. Use the molecular mass first, exactly as in Question 42. An alkali metal reacting with water gives the hydroxide $\mathrm{AOH}$, and if that has mass 40 then A has atomic mass $40 – 17 = 23$, which is sodium.

So $\mathrm{A} = \mathrm{Na}$ and $\mathrm{B} = \mathrm{NaOH}$:

$$2\mathrm{Na} + 2\mathrm{H_2O} \rightarrow 2\mathrm{NaOH} + \mathrm{H_2}$$

Now the second half. Aluminium oxide is amphoteric — Questions 31 and 46 — so it reacts with the alkali as though it were an acid, and the soluble product is sodium aluminate:

$$\mathrm{Al_2O_3} + 2\mathrm{NaOH} \rightarrow 2\mathrm{NaAlO_2} + \mathrm{H_2O}$$

so $\mathrm{C} = \mathrm{NaAlO_2}$.

This question and Question 42 have the same first half and different second halves, which is worth noticing: the exemplar is testing the molecular-mass route twice, once on its own and once as the setup for the amphoteric reaction.

Question 55

Give the reaction involved during extraction of zinc from its ore by

(a) roasting of zinc ore (b) calcination of zinc ore

Solution. The two words name two different first steps, applied to two different zinc ores, and the whole of the difference is whether air is present.

(a) Roasting is applied to the sulphide ore, zinc blende, and it is done by heating strongly in the presence of air:

$$2\mathrm{ZnS}(s) + 3\mathrm{O_2}(g) \xrightarrow{\text{Heat}} 2\mathrm{ZnO}(s) + 2\mathrm{SO_2}(g)$$

The oxygen of the air is a reactant, which is why the equation has $\mathrm{O_2}$ in it.

(b) Calcination is applied to the carbonate ore, calamine, and it is done by heating in the absence of air:

$$\mathrm{ZnCO_3}(s) \xrightarrow{\text{Heat}} \mathrm{ZnO}(s) + \mathrm{CO_2}(g)$$

There is no $\mathrm{O_2}$ in this equation, which is the visible signature of calcination.

Both routes arrive at the same place — zinc oxide — and the extraction then continues identically, by reducing that oxide with carbon:

$$\mathrm{ZnO} + \mathrm{C} \xrightarrow{\text{Heat}} \mathrm{Zn} + \mathrm{CO}$$

That convergence is the point of Question 39: the two preparations exist to reach an oxide, because an oxide is what carbon can reduce.

Question 56

A metal M does not liberate hydrogen from acids but reacts with oxygen to give a black colour product. Identify M and black coloured product and also explain the reaction of M with oxygen.

Solution. Does not liberate hydrogen from acids places M below hydrogen in the reactivity series, which narrows it to copper, mercury, silver, gold and platinum.

Reacts with oxygen to give a black product then picks out copper. Heating copper in air gives copper(II) oxide, which is black:

$$2\mathrm{Cu}(s) + \mathrm{O_2}(g) \xrightarrow{\text{Heat}} 2\mathrm{CuO}(s)$$

so $\mathrm{M} = \mathrm{Cu}$ and the black product is $\mathrm{CuO}$.

The visible change is striking and is the reason this reaction is used as a classroom demonstration: a bright red-brown copper wire, heated in a flame, acquires a dull black coating. That is the reaction referred to in Question 16 of the first unit, and it is a genuine chemical change rather than tarnishing.

Silver would have given black silver sulphide with hydrogen sulphide, but not with oxygen, so it does not fit. Gold and platinum react with oxygen not at all.

Note also that copper’s green coating in air — Question 16 of the second half — is a different product again, basic copper carbonate, formed slowly at room temperature rather than quickly on heating.

Question 57

An element forms an oxide $\mathrm{A_2O_3}$ which is acidic in nature. Identify A as a metal or non-metal.

Solution. A is a non-metal.

The rule is short and is worth stating as a rule, because the question is a one-step deduction from it:

• Metals form basic oxides — $\mathrm{Na_2O}$, $\mathrm{CaO}$, $\mathrm{MgO}$ all give bases with water and neutralise acids.
• Non-metals form acidic oxides — $\mathrm{CO_2}$, $\mathrm{SO_2}$, $\mathrm{P_2O_5}$ all give acids with water and neutralise bases.

The oxide here is stated to be acidic, so its element is a non-metal. The formula $\mathrm{A_2O_3}$ is consistent — a trivalent element of Group 15, such as phosphorus or nitrogen, gives exactly this formula.

One caution the syllabus expects you to know: the rule has exceptions in the amphoteric oxides, $\mathrm{Al_2O_3}$ and $\mathrm{ZnO}$, which are metal oxides that also behave as acids. But those are described as amphoteric, not as acidic, and the question says acidic — so the ordinary rule applies.

Question 58

A solution of $\mathrm{CuSO_4}$ was kept in an iron pot. After few days the iron pot was found to have a number of holes in it. Explain the reason in terms of reactivity. Write the equation of the reaction involved.

Solution. Iron is more reactive than copper, so it displaces copper from copper sulphate solution:

$$\mathrm{Fe}(s) + \mathrm{CuSO_4}(aq) \rightarrow \mathrm{FeSO_4}(aq) + \mathrm{Cu}(s)$$

Every place where the solution touched the pot, iron atoms went into solution as $\mathrm{Fe^{2+}}$ and copper came out onto the surface. The iron of the pot is therefore steadily eaten away, and after some days enough has dissolved for the pot to be perforated — the holes are literally where the iron used to be.

There are two visible signs that confirm it. The blue colour of the copper sulphate fades to the pale green of iron sulphate, and a red-brown deposit of copper appears on the inside of the pot.

The practical lesson is the one the question is really after: a solution must not be stored in a container made of a metal more reactive than the metal in the solution. It is the same reasoning as Question 44 of the first unit, where zinc chloride could not be kept in an aluminium container for exactly this reason.

Long Answer Questions

Question 59

A non-metal A which is the largest constituent of air, when heated with $\mathrm{H_2}$ in 1:3 ratio in the presence of catalyst (Fe) gives a gas B. On heating with $\mathrm{O_2}$ it gives an oxide C. If this oxide is passed into water in the presence of air it gives an acid D which acts as a strong oxidising agent.

(a) Identify A, B, C and D (b) To which group of periodic table does this non-metal belong?

Solution.

(a) The largest constituent of air is nitrogen, which makes up about 78 per cent of it, so $\mathrm{A} = \mathrm{N_2}$.

Nitrogen with hydrogen in a $1:3$ ratio over an iron catalyst is the Haber process, and it gives ammonia:

$$\mathrm{N_2} + 3\mathrm{H_2} \xrightarrow[\text{Fe}]{\text{Heat}} 2\mathrm{NH_3}$$

so $\mathrm{B} = \mathrm{NH_3}$. Note how precisely the $1:3$ ratio identifies the reaction — it is the stoichiometry of this equation and of no other.

Heating nitrogen with oxygen gives nitrogen monoxide:

$$\mathrm{N_2} + \mathrm{O_2} \rightarrow 2\mathrm{NO}$$

so $\mathrm{C} = \mathrm{NO}$.

Passing that oxide into water in the presence of air gives nitric acid — the air first oxidises the $\mathrm{NO}$ to $\mathrm{NO_2}$, which then dissolves:

$$4\mathrm{NO_2} + \mathrm{O_2} + 2\mathrm{H_2O} \rightarrow 4\mathrm{HNO_3}$$

so $\mathrm{D} = \mathrm{HNO_3}$, and the description strong oxidising agent confirms it — that is exactly the property behind Questions 7, 21 and 40.

(b) Nitrogen has five electrons in its outer shell and belongs to Group 15.

Question 60

Give the steps involved in the extraction of metals of low and medium reactivity from their respective sulphide ores.

Solution. Both start from a sulphide ore and both start with roasting, but they diverge immediately afterwards — and the reason is how tightly the metal holds its oxygen.

For a low reactivity metal such as mercury or copper, the oxide formed on roasting is so unstable that further heating decomposes it straight to the metal. No reducing agent is needed at all; this is what happens in Question 47.

For a medium reactivity metal such as zinc or iron, the oxide is stable and has to be reduced, normally with carbon, before the metal appears.

Both routes then finish with refining, usually electrolytic, to remove the remaining impurities.

Worked examples of each are $2\mathrm{HgS} + 3\mathrm{O_2} \rightarrow 2\mathrm{HgO} + 2\mathrm{SO_2}$ followed by $2\mathrm{HgO} \rightarrow 2\mathrm{Hg} + \mathrm{O_2}$ for the low-reactivity route, and $2\mathrm{ZnS} + 3\mathrm{O_2} \rightarrow 2\mathrm{ZnO} + 2\mathrm{SO_2}$ followed by $\mathrm{ZnO} + \mathrm{C} \rightarrow \mathrm{Zn} + \mathrm{CO}$ for the medium one.

Question 61

Explain the following

(a) Reactivity of Al decreases if it is dipped in $\mathrm{HNO_3}$ (b) Carbon cannot reduce the oxides of Na or Mg (c) NaCl is not a conductor of electricity in solid state whereas it does conduct electricity in aqueous solution as well as in molten state (d) Iron articles are galvanised. (e) Metals like Na, K, Ca and Mg are never found in their free state in nature.

Solution.

(a) Nitric acid is a strong oxidising agent, and on aluminium it forms a thin, tough, impervious layer of aluminium oxide, $\mathrm{Al_2O_3}$, over the surface. That layer seals the metal beneath it from any further attack, so the aluminium appears much less reactive than it is. The metal is said to be passivated — the same effect that lets dilute nitric acid be stored in aluminium containers, in Question 44 of the first unit.

(b) Carbon can only reduce the oxide of a metal that is less reactive than carbon itself. Sodium and magnesium are more reactive than carbon, so they hold their oxygen more strongly than carbon does and carbon cannot take it from them. Those metals have to be extracted by electrolysis of the molten salt instead, which is why they are in the top band of the extraction table.

(c) An ionic compound conducts only when its ions can move. In solid sodium chloride every ion is locked into a rigid crystal lattice and can only vibrate in place, so there is no charge transport and the solid does not conduct. Melting the salt or dissolving it in water breaks the lattice apart, the ions become free to migrate towards the electrodes, and the substance conducts well. This is the reasoning behind Question 10.

(d) Galvanising coats iron with zinc, and it protects in two ways. The zinc layer physically excludes the air and moisture that cause rusting; and because zinc is more reactive than iron, even a scratched coating goes on protecting, since the zinc corrodes in preference to the iron. That second mechanism is sacrificial protection, and it is why galvanising beats painting for outdoor ironwork.

(e) These four are highly reactive metals near the top of the reactivity series. Over geological time any free sodium, potassium, calcium or magnesium would have combined at once with the oxygen, water or carbon dioxide around it, so they survive only as compounds — chlorides, carbonates, sulphates and silicates. Only the least reactive metals, gold and silver, are found native, as Question 11 established.

Question 62

(i) Given below are the steps for extraction of copper from its ore. Write the reaction involved.

(a) Roasting of copper (I) sulphide (b) Reduction of copper (I) oxide with copper (I) sulphide. (c) Electrolytic refining

(ii) Draw a neat and well labelled diagram for electrolytic refining of copper

Solution.

(i)(a) Roasting. Copper(I) sulphide is heated strongly in air, and part of it is converted to the oxide:

$$2\mathrm{Cu_2S}(s) + 3\mathrm{O_2}(g) \xrightarrow{\text{Heat}} 2\mathrm{Cu_2O}(s) + 2\mathrm{SO_2}(g)$$

(i)(b) Auto-reduction. The oxide just made is then heated with the sulphide that remains. They reduce each other, and copper metal is set free with no external reducing agent:

$$2\mathrm{Cu_2O} + \mathrm{Cu_2S} \xrightarrow{\text{Heat}} 6\mathrm{Cu}(s) + \mathrm{SO_2}(g)$$

This is the reaction of Question 49(b), and it is why copper extraction needs no coke at this stage: the ore supplies its own reducing agent.

(i)(c) Electrolytic refining. The copper obtained is about 98 per cent pure and is refined electrolytically. The electrode reactions are:

$$\text{At the anode: } \mathrm{Cu}(s) \rightarrow \mathrm{Cu^{2+}}(aq) + 2e^-$$

$$\text{At the cathode: } \mathrm{Cu^{2+}}(aq) + 2e^- \rightarrow \mathrm{Cu}(s)$$

(ii) The cell has the impure copper as the anode, a strip of pure copper as the cathode, and acidified copper sulphate solution as the electrolyte. Copper dissolves off the anode, travels through the solution as $\mathrm{Cu^{2+}}$, and plates onto the cathode; the insoluble impurities fall to the floor of the vessel below the anode as anode mud.

Question 63

Of the three metals X, Y and Z. X reacts with cold water, Y with hot water and Z with steam only. Identify X, Y and Z and also arrange them in order of increasing reactivity.

Solution. The three conditions are the three bands from Question 4, and each names a group of metals.

X reacts with cold water — the most reactive band: potassium, sodium or calcium. Take sodium as the representative.

Y reacts with hot water only — this band has essentially one member in the syllabus, magnesium. (The key offers “Mg or Ca” here, but calcium reacts with cold water, as Question 6 established, so only magnesium actually fits the description.)

Z reacts with steam only — aluminium, zinc or iron. Take iron.

$$\mathrm{Na} : 2\mathrm{Na} + 2\mathrm{H_2O} \rightarrow 2\mathrm{NaOH} + \mathrm{H_2} \quad \text{(cold)}$$

$$\mathrm{Mg} : \mathrm{Mg} + 2\mathrm{H_2O} \rightarrow \mathrm{Mg(OH)_2} + \mathrm{H_2} \quad \text{(hot)}$$

$$\mathrm{Fe} : 3\mathrm{Fe} + 4\mathrm{H_2O} \rightarrow \mathrm{Fe_3O_4} + 4\mathrm{H_2} \quad \text{(steam)}$$

Increasing reactivity means least reactive first, and the harsher the condition a metal needs, the less reactive it is. Iron needs steam, magnesium only hot water, sodium only cold water, so:

$$\mathrm{Fe} < \mathrm{Mg} < \mathrm{Na}$$

A note on the printed key. It gives this as “Increasing reactivity series: Na > Mg > Fe”. That string is in decreasing order — it reads “sodium is more reactive than magnesium, which is more reactive than iron” — under a heading that says increasing. The chemistry is right and the label contradicts it; written as an increasing series it is $\mathrm{Fe} < \mathrm{Mg} < \mathrm{Na}$. Read the direction of the inequality signs rather than the word above them.

Question 64

An element A burns with golden flame in air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equations for the reactions involved.

Solution. Two independent clues fix the two elements.

Burns with a golden flame is the flame test for sodium, so $\mathrm{A} = \mathrm{Na}$. Atomic number 17 is chlorine, so $\mathrm{B} = \mathrm{Cl_2}$.

A metal and a non-metal give an ionic compound, and here it is common salt:

$$2\mathrm{Na} + \mathrm{Cl_2} \rightarrow 2\mathrm{NaCl}$$

so $\mathrm{C} = \mathrm{NaCl}$.

Electrolysing an aqueous solution of sodium chloride is the chlor-alkali process, and it gives sodium hydroxide together with chlorine and hydrogen:

$$2\mathrm{NaCl}(aq) + 2\mathrm{H_2O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{Cl_2}(g) + \mathrm{H_2}(g)$$

so $\mathrm{D} = \mathrm{NaOH}$, and the question’s mention of hydrogen being liberated confirms which electrolysis is meant.

The word aqueous is doing real work here. Electrolysing molten sodium chloride would give sodium metal and chlorine, with no hydrogen at all — that is the extraction of Question 18. With water present, hydrogen is discharged at the cathode in preference to sodium, and the sodium stays in solution as the hydroxide.

Question 65

Two ores A and B were taken. On heating ore A gives $\mathrm{CO_2}$ whereas, ore B gives $\mathrm{SO_2}$. What steps will you take to convert them into metals?

Solution. The gas each ore gives away identifies what kind of ore it is, and that in turn chooses the first step.

Ore A gives $\mathrm{CO_2}$ on heating, so it is a carbonate, $\mathrm{MCO_3}$. A carbonate is treated by calcination — heating in the absence of air:

$$\mathrm{MCO_3} \xrightarrow{\text{Calcination}} \mathrm{MO} + \mathrm{CO_2}$$

Ore B gives $\mathrm{SO_2}$, so it is a sulphide, $\mathrm{MS}$. A sulphide is treated by roasting — heating strongly in air:

$$2\mathrm{MS} + 3\mathrm{O_2} \xrightarrow{\text{Roasting}} 2\mathrm{MO} + 2\mathrm{SO_2}$$

Both preparations end at the same compound, the oxide $\mathrm{MO}$, which is exactly the point Question 39 makes: whatever the ore, the object is to reach an oxide, because an oxide is what can be reduced cheaply.

The second step is then the same for both. Reduce the oxide with carbon:

$$\mathrm{MO} + \mathrm{C} \xrightarrow{\text{Reduction}} \mathrm{M} + \mathrm{CO}$$

and finally refine the metal, usually electrolytically, to remove the last impurities.

So the full answer is two different first steps converging on one common second step — which is a good summary of the whole extraction section.

Common mistakes

• Swapping roasting and calcination, in Questions 39, 49, 55 and 65. Roasting is a sulphide heated in air; calcination is a carbonate heated without air. The giveaway in the equation is whether $\mathrm{O_2}$ appears among the reactants.
• Writing the metal oxide where the hydroxide belongs, in Questions 42 and 53. A metal reacting with water gives the hydroxide; the oxide comes from burning the metal in oxygen. The two are different reactions with different products.
• Carrying the diatomic subscript into the compound, in Question 48. $\mathrm{N_2}$, $\mathrm{O_2}$ and $\mathrm{Cl_2}$ describe the element as it exists; the formula of the compound comes from the ionic charges, so Li and $\mathrm{O_2}$ give $\mathrm{Li_2O}$ and not $\mathrm{LiO_2}$.
• Answering Question 44 without checking the direction. Gaining oxygen is oxidation and losing it is reduction. If both reactants come out going the same way, something has gone wrong — the two must always be opposite.
• Reversing an increasing series, as the printed key does in Question 63. “Increasing reactivity” means least reactive first: $\mathrm{Fe} < \mathrm{Mg} < \mathrm{Na}$. The key writes the decreasing order under the increasing label.
• Forgetting the word aqueous in Question 64. Electrolysing molten NaCl gives sodium metal; electrolysing the solution gives sodium hydroxide and hydrogen. One word changes the products entirely.
• Balancing $\mathrm{Al_2O_3} + \mathrm{NaOH}$ carelessly, in Questions 46 and 54. Two aluminiums need two $\mathrm{NaAlO_2}$, which need $2\mathrm{NaOH}$, giving exactly one water. It is the most-missed balance in the unit.
• Assuming carbon can reduce any oxide, in Question 61(b). Carbon only reduces the oxides of metals below carbon in the reactivity series. Sodium and magnesium are above it, which is why they must be extracted electrolytically.

Practise next

• Exemplar Multiple Choice Questions Q1–18 — physical properties and the water and acid reactions that Questions 42, 53 and 63 above depend on.
• Exemplar Multiple Choice Questions Q19–36 — ionic bonding, alloys and the refining cell, including the four-figure question that Question 62 asks you to draw from scratch.
• NCERT textbook Chapter 3 — Metals and Non-metals for the end-of-chapter exercise on extraction and the activity series.
• Textbook activities for the experiments behind Questions 37, 56 and 58 — the zinc-with-alkali test, the heated copper wire and the displacement in copper sulphate are all performed there.